Deriving the Gamma Function
Date: 12/15/2000 at 01:53:41 From: James Garcioparra Subject: Calculus/Probability/Statistics How can you prove that the square root of pi divided by 2 is 0.5 factorial? What is a fractional factorial like that equal to?
Date: 12/15/2000 at 19:02:14 From: Doctor Anthony Subject: Re: Calculus/Probability/Statistics Factorials can be generated by the Gamma function, and it is the Gamma function that allows us to define things like (1/2)! The Gamma function is defined as: 0 G(n) = INT [x^(n-1) e^(-x) dx] inf Now integrating by parts G(n) = [x^(n-1).(-e^(-x)] + INT[(n-1)x^(n-2).e^(-x).dx] The first bracket is zero at both ends when limits infinity and 0 are put in. So: G(n) = (n-1)INT[x^(n-2).e^(-x).dx] G(n) = (n-1).G(n-1) ........................................ and when n is a positive integer this defines the factorial function. G(n) = (n-1)(n-2)(n-3) ... 3 * 2 * 1 * G(1) and 0 G(1) = INT [e^(-x)] inf | 0 = -[e^(-x)] | | inf = -[0 - 1] = 1 It follows that: G(n) = (n-1)! It follows also that: G(1) = (1-1)! = 0! and we have seen that G(1) = 1, so then 0! = 1 . Now looking at non-integer values for the gamma function, we have for example: 0 G(1/2) = INT [x^(-1/2).e^(-x).dx] inf Putting x = u^2 and integrating this gives: 0 G(1/2) = 2.INT [e^(-u^2).du] inf This is the well-known integral for the normal distribution and we have: G(1/2) = sqrt(pi) Using this result we can now obtain all other positive half-integral values from the recurrence relation . For example: G(3/2) = (1/2).G(1/2) = sqrt(pi)/2 G(7/2) = (5/2).G(5/2) = (5/2)(3/2).G(3/2) = (15/4).sqrt(pi)/2 The recurrence relation  can also be used in defining the gamma function for negative values of n. Rewriting  as G(n) = G(n+1)/n, we have: G(-3/2) = G(-1/2)/(-3/2) = G(1/2)/[(-3/2)(-1/2)] = (4/3).sqrt(pi) For other values such as G(1/3) or whatever we have tables of values derived from integrating series approximations to the function. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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